Number of positive integral solutions of equation $\dfrac 1 x+ \dfrac 1 y= \dfrac 1 {n!}$

$$\dfrac 1 x+ \dfrac 1 y= \dfrac 1 {n!}$$

This is one of the popular equation to find out the number of solutions. From Google, here I found that for equation $\dfrac 1 x+ \dfrac 1 y= \dfrac 1 {n!}$, number of solutions are

$\psi(n)=\text{number of divisors of } n$.

$$\frac{\psi(n^2)+1}{2}$$

but when i turned here, it says that for equation $\dfrac 1 x+ \dfrac 1 y= \dfrac 1 {n!}$, number of solutions are $$\frac{\psi(n!^2)-1}{2}$$

However in the first link they explained the equation for $n=4$ and their formula correctly suits on that. But I want to confirm the correct answer.

If there exist any better way to get the number of positive integral solutions for the equation $\dfrac 1 x+ \dfrac 1 y= \dfrac 1 {n!}$, then please suggest me.

Would prime factorization help here? I have an algorithm to find out prime factors of any number. But not have any exact idea about implementing it over here.

My objective is to find out total number of positive integral solutions of the equation $\dfrac 1 x+ \dfrac 1 y= \dfrac 1 {n!}$.

Thank you.

If we let $N! = M$,

$$ \frac{1}{x}+\frac{1}{y} = \frac{1}{N!} = \frac{1}{M}$$

Let $x=M+a$ and $y=M+b$ where $a$ and $b$ are integers (positive or negative)

$$ \Rightarrow \frac{2M+a+b}{(M+a)(M+b)} = \frac{1}{M}$$

$$ 2M^2+M(a+b) = M^2+M(a+b)+ab$$

$$ \Rightarrow M^2 = ab$$

Now look at all the divisors of $M^2 = (N!)^2$ and find the values of $a$ and $b$ and hence find the values of $x$ and $y$.

It will be easier if you pick an example such as

$$ \frac{1}{x}+\frac{1}{y} = \frac{1}{3!} = \frac{1}{6} \tag{1}$$

and show that

$$\fbox{(7, 42), (8, 24), (9, 18), (10, 15), (12, 12), (15, 10), (18, 9), (24, 8),(42, 7),(5,-30),}\\ \fbox{(4,-12),(3,-6), (2, -3), (-3, 2), (-6, 3), (-12, 4), (-30, 5)} $$

are $(x,y)$ pairs that satisfy $(1)$, which when counted are $17$ pairs.

Hint: The number of divisors of $36=6^2 = d(6^2) = 9$, and they are $1,2,3,4,6,9,12,18, 36$.

What is the relationship between $d(6^2)$ and the number of solutions?

Note: I am just giving you an approach. It was not explicitly mentioned that $(x,y)$ pairs have to be positive. If that is the case, then you have to only consider positive pairs. (Henry rightly pointed out that it should be clear whether you are looking for only positive integer solutions).

Let's deal with the $N$ case before moving on to the $N!$ case

You have $\dfrac{1}{N} = \dfrac{1}{N+d} + \dfrac{1}{\frac{N^2}{d}+N}$ and you want all the denominators to be positive integers, which happens iff $d$ is a divisor of $N^2$. So the number of solutions is the number of divisors of $N^2$, but since almost all of these come in pairs, except when $d=N$, to de-duplicate you need to add $1$ and then divide by $2$ and your first link was correct.

Now moving on to the $N!$ case, if you write $N! = 2^{e_2}3^{e_3}5^{e_5}7^{e_7}11^{e_{11}}\cdots$ then the number of solutions is $\dfrac{1+ \prod_p (2e_p+1)}{2} $ where $e_p = \lfloor N/p^1 \rfloor + \lfloor N/p^2 \rfloor + \lfloor N/p^3 \rfloor + \cdots $, i.e.

$$\frac{1}{2} + \frac{1}{2} \prod_{p \text{ prime } \le N} \left(1+2 \sum_{i=1}^{\lfloor \log_p N \rfloor} \lfloor N/p^i \rfloor\right).$$

For example looking at $N!= 10!$, the answer would be $$[1 + (1+2(5+2+1))(1+2(3+1))(1+2(2))(1+2(1))]/2$$ $$ =[1+ 17 \times 9 \times 5 \times 3]/2 $$ $$= 1148$$